∫(lnx)² dx=?c= x(lnx)²-2xlnx+2x+c
integral
∫6x+10/(x+2)² dx ?c=ln(x+2)üssü6 +2/x+2+c
basit kesirlere ayırma yöntemi kullandım
6x+10/(x+2)2=A/(x+2) + B/(x+2)2
6x+10=Ax+2A+B
A=6 B=-2
∫[6/x+2 - 2/(x+2)2]dx
∫ 6/x+2 - 2∫1/(x+2)2dx
x+2=u dx=du
6ln|x+2|-2∫1/u2du
6ln|x+2| + 2/u +C
ln|x+2|6 + 2/x+2 + C
umarım yararlı olmuştur
6x+10/(x+2)2=A/(x+2) + B/(x+2)2
6x+10=Ax+2A+B
A=6 B=-2
∫[6/x+2 - 2/(x+2)2]dx
∫ 6/x+2 - 2∫1/(x+2)2dx
x+2=u dx=du
6ln|x+2|-2∫1/u2du
6ln|x+2| + 2/u +C
ln|x+2|6 + 2/x+2 + C
umarım yararlı olmuştur
Süleyman Oymak 16:24 10 May 2013 #4 